Problem: Let $g(x)=x^5+x^4$. On which intervals is $g$ decreasing? Choose 1 answer: Choose 1 answer: (Choice A) A $x<-\dfrac45$ only (Choice B) B $-\dfrac45<x<0$ only (Choice C) C $x<-\dfrac45$ and $x>0$ (Choice D) D $x<0$ only (Choice E) E The entire domain of $g$
We can analyze the intervals where $g$ is increasing/decreasing by looking for the intervals where its derivative $g'$ is positive/negative. A function can only change its direction from increasing to decreasing and vice versa between its critical points and the points where the function itself is undefined. The derivative of $g$ is $g'(x)=x^3(5x+4)$. $g'(x)=0$ for $x=0,-\dfrac45$. Since $g'$ is a polynomial, it's defined for all real numbers. Therefore, our critical points are $x=0$ and $x=-\dfrac45$. $g$ is defined for all real numbers so we only need to consider the critical points. Our critical points divide the number line into three intervals: $\llap{-}3$ $\llap{-}2$ $\llap{-}1$ $0$ $1$ $2$ $3$ $ x<-\frac{4}{5}$ $ -\frac{4}{5}<x<0$ $ -\frac{4}{5}$ $x>0$ Let's evaluate $g'$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $g'(x)$ Verdict $x<-\dfrac45$ $x=-1$ $g'(-1)=1>0$ $g$ is increasing $\nearrow$ $-\dfrac45<x<0$ $x=-\dfrac{1}{5}$ $g'\left(-\dfrac{1}{5}\right)=-\dfrac{3}{125}<0$ $g$ is decreasing $\searrow$ $x>0$ $x=1$ $g'(1)=9>0$ $g$ is increasing $\nearrow$ In conclusion, $g$ is decreasing over the interval $-\dfrac45<x<0$ only.